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Question:

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

2GmM/3R

GmM/2R

GmM/3R

5GmM/6R

Solution:

Given that,
Mass of satellite = m
Mass of planet = M
Radius = R
Altitude h = 2R
Now, The gravitational potential energy P.E = -GMm/r
Potential energy at altitude = -GMm/(3R)
Orbital velocity v₀ = √(2GM/(R+h))
Now, the total energy is
Ef = (1/2)mv₀² - GMm/(3R)
Ef = (1/2)GMm/(3R) - GMm/(3R)
Ef = GMm/(3R)[(1/2) - 1]
Ef = -GMm/(6R)
Now, Ei = Ef
Now, the minimum required energy
K.E = GMm/R - GMm/(6R)
K.E = 5GMm/(6R)
Hence, the minimum required energy is 5GMm/(6R)